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PostPosted: Sun Feb 05, 2012 20:49 
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dcbwhaley wrote:
[Energy is always conserved.] Momentum isn't.

Pardon?
Could you explain that, please?


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PostPosted: Mon Feb 06, 2012 10:02 
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Steve wrote:
dcbwhaley wrote:
[Energy is always conserved.] Momentum isn't.

Pardon?
Could you explain that, please?


Certainly. And to a man who says "I know physics - quite well too"

The law of conservation of energy istates that the total amount of energy in an isolated system remains constant over time. The total energy is said to be conserved over time. For an isolated system, this law means that energy can change its location within the system, and that it can change form within the system, for instance chemical energy can become kinetic energy, but that energy can be neither created nor destroyed. So in accelerating a car the chemical energy in the tank is converted to kinetic energy and into heat in the engine. And in braking the kinetic energy is converted to heat in the discs.

The law of conservation of linear momentum states that if no external force acts on a closed system of objects, the momentum of the closed system remains constant. But if an external force is applied momentum is no longer conserved. If that were not so it would be impossible to stop your car without making another object accelerate and road junctions would be very interesting places :D

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PostPosted: Mon Feb 06, 2012 13:55 
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dcbwhaley wrote:
Certainly. And to a man who says "I know physics - quite well too"

Let me prove that!

I have no issue with the following:

dcbwhaley wrote:
The law of conservation of energy istates that the total amount of energy in an isolated system remains constant over time. The total energy is said to be conserved over time. For an isolated system, this law means that energy can change its location within the system, and that it can change form within the system, for instance chemical energy can become kinetic energy, but that energy can be neither created nor destroyed. So in accelerating a car the chemical energy in the tank is converted to kinetic energy and into heat in the engine. And in braking the kinetic energy is converted to heat in the discs.

However:
dcbwhaley wrote:
The law of conservation of linear momentum states that if no external force acts on a closed system of objects, the momentum of the closed system remains constant. But if an external force is applied momentum is no longer conserved. If that were not so it would be impossible to stop your car without making another object accelerate and road junctions would be very interesting places :D

As correct as your quote (the part in in italics) is, you have added an interpretation which isn’t correct, and isn’t a physical parallel.

If external energy is applied to a system, then the system isn’t isolated; otherwise the energy is conserved.
Likewise, if an external force is applied to a system, then the system isn’t isolated; otherwise the momentum is conserved.

Dave, have you never hear of the fundamental physical low of “conservation of momentum”? I thought this was CGSE stuff!

Oh, and it is perfectly possible to “stop your car” and have the momentum conserved, you just don’t “see” what is going on (the lack of seeing shouldn’t prevent you from understanding it). What actually happens is that the junction really is accelerating, but due to the relative mass of our planet it is moving by such a small amount that we can't measure it. Proving my point: consider our planet instead being a 10m diameter plastic ball. Assuming a vehicle had traction with the plastic surface, and used the ball to decelerate, can you really say that the plastic road surface wouldn’t accelerate at all?

The momentum is always conserved for the isolated car/planet system.
Either that or you have just found an entirely way to travel though space: a 'non-reaction engine'!

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PostPosted: Mon Feb 06, 2012 21:38 
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I am quite puzzled, Steve, but wish to learn from your great knowledge of physics.

In a game of snooker the player strikes the white ball with his cue and it moves across the baize with some velocity. But the player and the cue continue in the same direction as he follows through. What object moves backwards to conserve the momentum? And it turns out to be a foul shot - the cue ball goes round three cushions before coming to a halt - where has its momentum been conserved?

And I could ask the same about cricket - where a cover drive reverses the momentum of the ball without imparting any momentum to the bat or the batsman. Or about any racket sport.

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PostPosted: Mon Feb 06, 2012 21:58 
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dcbwhaley wrote:
I am quite puzzled, Steve, but wish to learn from your great knowledge of physics.

In a game of snooker the player strikes the white ball with his cue and it moves across the baize with some velocity. But the player and the cue continue in the same direction as he follows through. What object moves backwards to conserve the momentum? And it turns out to be a foul shot - the cue ball goes round three cushions before coming to a halt - where has its momentum been conserved?

And I could ask the same about cricket - where a cover drive reverses the momentum of the ball without imparting any momentum to the bat or the batsman. Or about any racket sport.


And apart from talking a lot of round sperical objects - the purpose of the above is .............................. :?: :?:

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PostPosted: Mon Feb 06, 2012 22:39 
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dcbwhaley wrote:
I am quite puzzled, Steve, but wish to learn from your great knowledge of physics.

In a game of snooker the player strikes the white ball with his cue and it moves across the baize with some velocity. But the player and the cue continue in the same direction as he follows through. What object moves backwards to conserve the momentum? And it turns out to be a foul shot - the cue ball goes round three cushions before coming to a halt - where has its momentum been conserved?

And I could ask the same about cricket - where a cover drive reverses the momentum of the ball without imparting any momentum to the bat or the batsman. Or about any racket sport.


I hope you are not questioning one of the FUNDAMENTAL LAWS OF PHYSICS!

Had you attempted you address my prior example to you, I think you would have been quickly enlightened (by simply incrementing the scale). You really need to think about that particular example before you can properly understand my answers to the example you gave (below), otherwise you will never understand this critical concept and the whole concept of "isolated" and "external" will be forever lost on you.


Cricket
Ball weight = 160g. Professional level speed delta (bat+bowl) = 200mph = 90m/s. ρ=14.4kgm/s. Equivalent to 80kg batsman moving forward at 0.18m/s, easily done by rocking forward on feet - which is exactly what cricketers do! The force is transferred through their feet; given the non-zero foot base (like a car's wheelbase but with feet instead), force can be transferred without causing unbalance. The ground provides the reactive force and accelerates appropriately (remember, your inability to see something immeasurably small doesn't mean it isn't there).

Snooker
Ball weight = 150g. Typical "break" speed 24mph = 11m/s. ρ=1.65kgm/s. Equivalent to 80kg player moving forward at 0.02m/s - practically imperceptible, especially when the player usually stands afterwards anyway.

The able sides provides the reactive force to accelerate ball; the impact force is transferred through the table legs to the ground (and so on); they tend not to move much, unless you hit it hard with a sledgehammer!
The ball rolling on the felt is subject to friction, the felt providing the reactive force for deceleration, transferred via glue to table top, transferred through legs to ground (and so on).
Had the table been on an ideal frictionless surface (i.e. no way to transfer the forces to the ground), you would just about be able to see the table moving back and forth (the 'rest state' being dependent on how the player was in contact with the table); if the table was light, that movement would be much more obvious.

Check any and all of that with any professionals who regularly exercise their physics qualifications in their day job.

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PostPosted: Mon Feb 06, 2012 23:25 
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Steve wrote:
I hope you are not questioning one of the FUNDAMENTAL LAWS OF PHYSICS!

No Steve. I am not questioning a fundamental law of Physics which states if no external force acts on a closed system of objects, the momentum of the closed system remains constant

Quote:
Cricket
Ball weight = 160g. Professional level speed delta (bat+bowl) = 200mph = 90m/s. ρ=14.4kgm/s. Equivalent to 80kg batsman moving forward at 0.18m/s, easily done by rocking forward on feet - which is exactly what cricketers do! The force is transferred through their feet; given the non-zero foot base (like a car's wheelbase but with feet instead), force can be transferred without causing unbalance. The ground provides the reactive force and accelerates appropriately (remember, your inability to see something immeasurably small doesn't mean it isn't there).


When an old cricketer, before leaving the crease, makes a forward defensive stroke at a ball from the pavilion end three things happen: the ball reverses direction towards the pavilion; the bat moves in the direction of the pavilion: the old cricketer moves towards the pavilion. Nothing moves towards the gasworks end to conserve the momentum. That is because an external force is being applied which means momentum is not conserved.

Quote:
Snooker
Ball weight = 150g. Typical "break" speed 24mph = 11m/s. ρ=1.65kgm/s. Equivalent to 80kg player moving forward at 0.02m/s - practically imperceptible, especially when the player usually stands afterwards anyway.

Again all movement is forward. No reverse momentum. External force means momentum is not conserved.

Quote:
Check any and all of that with any professionals who regularly exercise their physics qualifications in their day job.
Done that. Until last April I was earning £60K per annum through applying my Physics degree to my day job

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PostPosted: Mon Feb 06, 2012 23:36 
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I’ve just realised that I probably should proceed with my "plastic ball" example.

Some questions that you should ask yourself, Dave:

Assume a car was relatively stationary, on a 10m diameter solid plastic ball that was absolutely and rotationally stationary, and always had traction with the surface (all in the vaccum of outer space, and without any effect of gravity) - no external forces; an isolated system.
If the car then accelerates (say to a relative [surface] speed of 100mph), will the planet remain without rotation, or will it start rotating with the surface speed of a few mph?
Answer: it will start to spin with a some surface speed (what is to stop it?)

The volume of a sphere increases with the cube of the length of one of the diameter, thus so would the weight.

If the ball was instead 100m in diameter, would it still spin?
Answer: yes but at a much lower rotational rate, probably at one thousandth of the original surface speed

If the ball was instead 1000m (1km) in diameter, would it still spin?
Answer: yes but at a much lower rotational rate, probably at one millionth (1x10E6) of the original surface speed

If the ball was instead 10,000m (10km) in diameter …
(1x10E9)

If the ball was instead 100,000m (100km) in diameter …
(1x10E12)

If the ball was instead 1,000,000m (1000km) in diameter …
(1x10E15)

If the ball was instead 12,000,000m (12000km) in diameter (which it is) …
(<1x10E18)


You should be able to understand that the car still transfers the force and still results with, although vanishing small, some acceleration to the planet.
Momentum is always conserved!

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PostPosted: Mon Feb 06, 2012 23:47 
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Steve wrote:
I’ve just realised that I probably should proceed with my "plastic ball" example.

Some questions that you should ask yourself, Dave:

Assume a car was relatively stationary, on a 10m diameter solid plastic ball that was absolutely and rotationally stationary, and always had traction with the surface (all in the vaccum of outer space, and without any effect of gravity) - no external forces; an isolated system.
If the car then accelerates (say to a relative [surface] speed of 100mph), will the planet remain without rotation, or will it start rotating with the surface speed of a few mph?
Answer: it will start to spin with a some surface speed (what is to stop it?)

...

You should be able to understand that the car still transfers the force and still results with, although vanishing small, some acceleration to the planet.


Of course. In those circumstances with no external force (i.e force from outside the closed system) applied momentum will be conserved. But if a rocket from Ursa Major were to smash into the side of the car it would move rapidly in the direction of Polaris with no corresponding movement in the opposite direction

Quote:
Momentum is always conserved!


Are you claiming that momentum is conserved in ac losed system even when an external force is applied?

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PostPosted: Mon Feb 06, 2012 23:48 
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dcbwhaley wrote:
No Steve. I am not questioning a fundamental law of Physics which states if no external force acts on a closed system of objects, the momentum of the closed system remains constant

It seems you are failing to understand the concept of a "closed system".

dcbwhaley wrote:
When an old cricketer, before leaving the crease, makes a forward defensive stroke at a ball from the pavilion end three things happen: the ball reverses direction towards the pavilion; the bat moves in the direction of the pavilion: the old cricketer moves towards the pavilion. Nothing moves towards the gasworks end to conserve the momentum. That is because an external force is being applied which means momentum is not conserved.

You need to explain how all the forces are resolved and what their reactions are (I thought this was obvious); this should be very easy for you to do.

dcbwhaley wrote:
Again all movement is forward. No reverse momentum. External force means momentum is not conserved.

Why must the player be considered as "external"?
Dave, please do explain to the reader exactly what happens if the player (and planet) is considered within the bounds of a "closed system".

Ironically, say yet again, you need to be able to understand my example of the ball; and don't forget the INITIAL CONDITIONS (hint)

dcbwhaley wrote:
Quote:
Check any and all of that with any professionals who regularly exercise their physics qualifications in their day job.
Done that. Until last April I was earning £60K per annum through applying my Physics degree to my day job

Then you are about to be "schooled" in your own specialist subject!

Momentum is always conserved, your problem is that you don't know how to "close" the system. Prove me wrong!

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PostPosted: Tue Feb 07, 2012 00:05 
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dcbwhaley wrote:
But if a rocket from Ursa Major were to smash into the side of the car it would move rapidly in the direction of Polaris with no corresponding movement in the opposite direction

How utterly stooopid!

There is acceleration of some component of the system towards Ursa Major (away from what I assume is Polaris for your example); the problem here is that you don't understand what it is.
The car accelerates towards Polaris; the rocket accelerates towards Ursa Major. So no problems there!

For your latest example: ρ [of rocket] + ρ [of car and planet it is on] before the collision is still the same as ρ [of rocket + car + planet]. So how can you prove this wrong?

Let's reduce your example to prove the silliness of your claim.
Initial conditions: a snooker ball (2) from Ursa Major travelling at 1010m/s, heads straight towards another identical snooker ball (1) again from Ursa Major travelling at 1000m/s(as seen relative to an observer moving at 1005m/s in the direction of Ursa Major)
Event: the two collide and (for argument's sake) meld with each other.
Rest state: the combined speed will be 1005m/s.
ρ(1) + ρ(2) = ρ(1+2). Momentum is conserved!
(1) accelerated by 5m/s away from Ursa Major; (2) accelerated by 5m/s towards Ursa Major - there's your "corresponding movement in the opposite direction". Momentum of the system is still conserved!
What is your problem with that, Dave!!

If you want to expand it wider to consider variables you will invariably will claim that disprove the conservation of momentum: the net momentum of: whatever the rocket took off from, the rocket gases, the rocket, the car (and it's subsequent fragments) (and the planet it is on), is conserved, from initial conditions to rest state.
System closed; job done!

If you can't see how momentum is conserved, then your model is somehow wrong (be it incomplete).
How can someone of your claimed calibre not understand this simple and fundamental concept?


dcbwhaley wrote:
Are you claiming that momentum is conserved in ac losed system even when an external force is applied?

Misrepresentation! Ever the clue of a failed argument, or trolling…
Steve previously wrote:
if an external force is applied to a system, then the system isn’t isolated; otherwise the momentum is conserved.


Moreover, I made the same parallel with energy:
Steve previously wrote:
If external energy is applied to a system, then the system isn’t isolated; otherwise the energy is conserved.
...which, rather strangely, you didn't dispute; yet you still consider the two attributes to be different in this respect.

So Dave, tell us your claimed difference between the "conservation of momentum" and the "conservation of energy"; how does one fail where the other succeeds?
By your logic, you seem to be claiming that external energy can be introduced into an otherwise closed system and that energy is still conserved; can you clarify this. If that’s not the case, then please do explain the difference between that and what you are saying with externally affected momentum.

I would love to see you try to reconcile that one!





Last, but certainly not least:
dcbwhaley wrote:
Of course. In those circumstances with no external force (i.e force from outside the closed system) applied momentum will be conserved.

You just accepted my original point !?!

You've just accepted there need not be an external force for the accelerating car example - you've just agreed that momentum is actually conserved for your same example explaining otherwise :lol:

You've just inherently shot yourself in the foot :lol:

System closed; argument closed!
Can we move on?

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PostPosted: Wed Feb 08, 2012 11:10 
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Probably too late for all this, but I've a vague recollection from O-level physics that momentum is only conserved in purely elastic collisions? The snooker balls were always cited as good examples of this, M1V1 = M2V2 & all that? If the collision is not purely elastic, energy is always conserved, but momentum is not. (2 cars coliding, M1, V1 is most certainly NOT M2V2)! The batsman is viscous. So, to an extent, are the bat and ball. Not really being a cricket many myself (or, indeed a physicist, for that matter!) it seems to me that between the ball and its contact point with the bat, a fair amount of momentum is conserved. As the bat flexes, less of it will be conserved at the handle of the bat - less still at the wrist of the batsman, and less again at the shoulder. Not really sure where this argument is going though!

It all reminds me of a question from many years ago... not sure I've got it quite right, but the gist is:

A train is travelling at (say)150km/k and a fly, travelling towards it at 0.1m/s hits the windscreen. The fly's velocity decreases,for an instant in time, then fly is stationary, and then it's velocity is 150km/h in the opposite direction. Is the train's velocity zero at the same time as the fly's velocity is zero, and if not, why not?


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PostPosted: Wed Feb 08, 2012 12:28 
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Mole wrote:
Probably too late for all this, but I've a vague recollection from O-level physics that momentum is only conserved in purely elastic collisions? The snooker balls were always cited as good examples of this, M1V1 = M2V2 & all that?

I know the answer and the associated subtleties, but before I give it I would really like to know DCB's understanding of this.

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PostPosted: Wed Feb 08, 2012 17:38 
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Mole wrote:
Probably too late for all this, but I've a vague recollection from O-level physics that momentum is only conserved in purely elastic collisions? The snooker balls were always cited as good examples of this, M1V1 = M2V2 & all that? If the collision is not purely elastic, energy is always conserved, but momentum is not. (2 cars coliding, M1, V1 is most certainly NOT M2V2)! The batsman is viscous. So, to an extent, are the bat and ball. Not really being a cricket many myself (or, indeed a physicist, for that matter!) it seems to me that between the ball and its contact point with the bat, a fair amount of momentum is conserved. As the bat flexes, less of it will be conserved at the handle of the bat - less still at the wrist of the batsman, and less again at the shoulder. Not really sure where this argument is going though!

It all reminds me of a question from many years ago... not sure I've got it quite right, but the gist is:

A train is travelling at (say)150km/k and a fly, travelling towards it at 0.1m/s hits the windscreen. The fly's velocity decreases,for an instant in time, then fly is stationary, and then it's velocity is 150km/h in the opposite direction. Is the train's velocity zero at the same time as the fly's velocity is zero, and if not, why not?


Yes, I've heard that one before, and it's all quite clear what happens.

The bit of the train where the fly makes contact does indeed stop briefly, before accelerating up to more or less its original speed. I reckon this is because of very localised deflection and distortion of the front of the train, although the exact nature of such deflections and distortion are naturally governed by the precise nature of the crumple zones on the unfortunate fly - who will need a bit of T cut to sort him out.

At any rate the train as a whole clearly doesn't stop; or if it did, where did all the poke come from to get it back up to speed in the proverbial twinkling of an eye?

The fly obviously comes to a brief stop, after suffering enormous G forces as it decellerates, and then it accelerates a bit quick afterwards to match the speed of the train which is now not travelling quite as fast as it was before the encounter with the fly. Anyhow, I trust we can accept that the train will have recovered most of its original speed.

The only thing to really lose out badly here seems to be the fly: it sure is a bit tough on him. However, if it's any comfort to anybody, the train should reveal no more than minor damage.

Best wishes all,
Dave.


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PostPosted: Wed Feb 08, 2012 18:58 
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Yes, I think that's it. Also, "the fly", I think, ceases to be a single body. So the first bit of fly to hit the train will (almost at a molecular level) deflect the nose of the train, then the next bit of fly, then the next, and so on. A bit like the cricket ball and the batsman, I think, in that while there is a fair bit of energy stored in the ball, a lot of it is absorbed by flexure of the bat, the "viscous" nature of the batsman's arms, and the general "inertia" of the whole system so that there's very little of that initial energy left to know the batsman over.


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PostPosted: Wed Feb 08, 2012 19:46 
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Steve wrote:
System closed; argument closed! Can we move on?


Since you are having to resort to abuse to bolster your arguments it would, indeed, be better for us to move on.

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PostPosted: Wed Feb 08, 2012 20:13 
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dcbwhaley wrote:
Steve wrote:
System closed; argument closed! Can we move on?


Since you are having to resort to abuse to bolster your arguments it would, indeed, be better for us to move on.

What abuse, exactly?
How did that (whatever it was) compare to what is now apparent to have been your earlier sarcasm?
How did it (whatever it was) bolster anything?

A person who wanted to mask their realisation that they were outright wrong, would claim much the same. I think it is fairly obvious what occurred here.

I'm very happy to carry on if you still support your original claim that I asked you to explain.

:hello:

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PostPosted: Wed Feb 08, 2012 20:24 
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TripleS wrote:
Mole wrote:
A train is travelling at (say)150km/k and a fly, travelling towards it at 0.1m/s hits the windscreen. The fly's velocity decreases,for an instant in time, then fly is stationary, and then it's velocity is 150km/h in the opposite direction. Is the train's velocity zero at the same time as the fly's velocity is zero, and if not, why not?


Yes, I've heard that one before, and it's all quite clear what happens.

The bit of the train where the fly makes contact does indeed stop briefly, before accelerating up to more or less its original speed. I reckon this is because of very localised deflection and distortion of the front of the train, although the exact nature of such deflections and distortion are naturally governed by the precise nature of the crumple zones on the unfortunate fly - who will need a bit of T cut to sort him out.

There will be a force from the impact that will quickly propagate through both the train and the fly; this propagation is not instantaneous. The point to note here is that one part of the fly (being compressible/compactable matter) may not be travelling at the same speed as other parts of the fly (the same applies for the train to some extent).

However, basic logic suggests that if the two objects in this case were absolutely identical and the collision was purely inelastic (only deformation), no part of the mass doing 150kph will ever fall below 75kph (or 74.95kph) [think of the two coming together at + and - 75kph, then add +75kph to both].
No part of the mass doing the 150kph would ever end up at the 0kph, unless the collision was elastic enough to propel some part of the other mass to 300kph at some point.

It gets a bit more complicated when considering the different atomic makeups of the fly and train, so I will stop there!

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PostPosted: Wed Feb 08, 2012 21:02 
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Mole wrote:
Probably too late for all this, but I've a vague recollection from O-level physics that momentum is only conserved in purely elastic collisions? The snooker balls were always cited as good examples of this, M1V1 = M2V2 & all that? If the collision is not purely elastic, energy is always conserved, but momentum is not. (2 cars coliding, M1, V1 is most certainly NOT M2V2)!

Actually, the momentum is always conserved, regardless of elasticity.

I will try my own explanation of this, using the basic physical equation F=ma and knowledge that every force has an equal and opposite 'reactive' force. Hopefully this is clear enough to be understood.

So the force acting on two colliding objects: F(1) = -F(2) (the minus is needed to account for the opposing direction of the force)

Substitute F=ma, so ma(1) = -ma(2)

Also, V = at
[change of] velocity = acceleration x time (from v = u + a*Delta t, where u is initial velocity).

So mV/t(1) = -mV/t(2)

t is the same for both (obviously), so the mV on the left and right must cancel when summed, meaning no change of net mV.
"mV" is momentum - it is conserved.

An inelastic collision doesn't change the differential force between the two objects (although it might be spread out less/more over time [still equal for both objects]), so the principle remains valid and momentum is conserved regardless.
If anything isn't conserved, it is kinetic energy.

If the mass of one was (say) halved, then the acceleration of that mass must also have relatively doubled (F=ma) and so the (change of) velocity would also have doubled. mV is still conserved.

In the case of the elastic collision, none of the kinetic energy is transferred to another form, (which allows imparting some velocity in say a rotational form).
In the case of an inelastic collision, some of kinetic energy is transferred to another form, usually by means of deformation of the objects and heat.

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PostPosted: Wed Feb 08, 2012 21:10 
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Steve wrote:
I know the answer and the associated subtleties, but before I give it I would really like to know DCB's understanding of this.


The momentum of the fly + train system is conserved (since no external force is applied) so the train slows down slightly. Since the collision is inelastic kinetic energy isn't conserved some of it been converted to heat. Since the collision is inelastic - the fly is squashed - it isn't really meaningful to speak of the "velocity of the fly" during the collision as a single figure - you have to consider each portion of it separately. Whilst the net velocity might be zero that is because the part in contact with the train is travelling at the velocity of the train while the more distant part is still travelling in the opposite direction. That is why the train is never at zero velocity

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